# Given x1,x2 the roots of x^2+ax+1=0 and a integer, show that x1^k+x2^k and a^2 -1 are primes for k integer.

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You should come up with the following substitution such that:

`u_k = x_1^k + x_2^k`

Using Vieta's relation yields:

`x_1*x_2 = 1`

You may also use the following notation such that:

`u_(-k) = x_1^(-k) + x_2^(-k)`

`u_(-k) = 1/(x_1^k)+ 1/(x_2^k)`

`u_(-k) = (x_1^k + x_2^k)/(x_1^k*x_2^k)`

`u_(-k) = (x_1^k + x_2^k)/((x_1*x_2)^k)`

But `x_1*x_2 = 1 =gt u_(-k) = (x_1^k + x_2^k) = u_k`

Since `u_(-k) = u_k` , you may assume that k in N.

`x_1^k + x_2^k = (x_1 + x_2)*(x_1^(k-1) + x_2^(k-1)) - x_1*x_2*(x_1^(k-2) + x_2^(k-2))`

Using Vieta's relation yields:

`x_1 + x_2 = -a`

Hence, `u_k = x_1^k + x_2^k = -a*u_(k-1) - u_(k-2)`

Since `u_0 = x_1^0 + x_2^0 = 1+1 = 2` and `u_1 = -a` , yields that `u_k` is an integer number.

You need to prove that `u_k` and `a^2 - 1` are primes, such that:

`u_0 = 2 ; u_1 = -a` , using the relation `u_k = -a*u_(k-1) - u_(k-2) =gt u_2 = -a*(-a) - 2 =gt u_2 = a^2 - 2`

**Since the numbers `u_0, u_1, u_2` and `a^2 - 1` are primes, hence, `u_k` and `a^2 - 1` are also primes.**